Question

Giải hệ pt:

\(\left\{{}\begin{matrix}\frac{x+y}{3}+\frac{2}{3}=3\\\frac{4x-y}{6}+\frac{x}{4}=1\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{3}+\frac{y}{3}=\frac{7}{3}\\\frac{11}{12}x-\frac{y}{6}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{6}+\frac{y}{6}=\frac{7}{6}\left(1\right)\\\frac{11x}{12}-\frac{y}{6}=1\left(2\right)\end{matrix}\right.\)

Lấy (1) cộng (2) ta được

\(\frac{13}{12}x=\frac{13}{6}\Rightarrow x=2\Rightarrow y=5\)

Vậy.............

Answer 2

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=7\\8x-2y+3x=12\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13x=26\\y=7-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y=14\\11x-2y=12\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)

Vậy ....